| E. J. Brooksmith - 1889
...and produced to meet in C: prove that AC and BC are bisected at E and D. 10. Define compound ratio. **Equiangular parallelograms have to one another the ratio which is compounded of the** ratio of their sides. 1 1 . The rectangle contained by the diagonals of a quadrilateral figure inscribed... | |
| Euclid - 1890 - 400 sider
...CD = P : Q, = X:Y, = dupl. ratio of LM to NO. .-. AB : CD = LM : NO. 272 Proposition 23. THEOREM — **Equiangular parallelograms have to one another the ratio which is compounded of the ratios of** their sides. Let ABCD, CEFG be equiang. Os, in which AA BCD = EGG. Place them so that a pair of the... | |
| Edward Mann Langley, W. Seys Phillips - 1890 - 515 sider
...student to enunciate generally the proposition assumed, and to demonstrate it. PROPOSITION 23. THEOREM. **Equiangular parallelograms have to one another the ratio which is compounded of the ratios of** their sides. Let AC, CF be equiangr. ||gms such that L BCD= L ECG ; then ||gm AC : jgm CF in the ratio... | |
| Queensland. Department of Public Instruction - 1890
...the external bisector ? 8. Triangles which have one angle of the one equal to one angle of the other, **have to one another the ratio which is compounded of the ratios of the sides** about the equal angles. 9. The three external bisectors of the angles of a triangle cut the sides in... | |
| 1891
...diagonal is parallel to a side. 3. The areas of parallelograms which are equiangular to one another **have to one another the ratio which is compounded of the ratios of** their sides. Hence deduce that the areas of similar parallelograms are to one another in the duplicate... | |
| 1898
...externally their common tangent is a mean proportional between their diameters (4 marks). 8. Prove that **equiangular parallelograms have to one another the ratio which is compounded of the ratios of** their sides (10 marks). Show also that triangles which have one angle of the one equal or supplemental... | |
| Edinburgh Mathematical Society - 1899
...:NH =(EF : GH)2 But KAB:LCD= MF : NH (AB : CD)2 = (EF : GH)2 AB :CD = EF : GH EUCLID VI. 23. Mutually **equiangular parallelograms have to one another the ratio which is compounded of the ratios of** their sides. Let parallelogram BE be equiangular to parallelogram CD, and let _ to prove / / / .|pBE:||-CD... | |
| Eldred John Brooksmith - 1901
...on the diagonal of the rectangle. 11. Prove that parallelograms which are equiangular to one another **have to one another the ratio which is compounded of the ratios of** their sides. 12. Prove that, in any right-angled triangle, any rectilineal figure described on the... | |
| 1903
...together equal to two right angles. (c) Shew that parallelograms which are equiangular to one another **have to one another the ratio which is compounded of the ratios of** their sides. 3. Describe a circle which shall pass through a given point and touch two given straight... | |
| Euclid - 1904 - 456 sider
...magnitude. [Book v. Def. 11.] PROPOSITION 23. THEOREM. Parallelograms which are equiangular to one another **have to one another the ratio which is compounded of the ratios of** their sides. ADH \ KLMEF Let the parm AC be equiangular to the parm CF, having the L. BCD equal to... | |
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